BOF's probability problem...
BOF's probability problem...
Dear All,
many years ago (quite a big many), I set a probability teaser in one of the early BDGA newsletters.
I cannot find any copy of my original working, either on computers or on paper!
If anyone has an archive of my original problem and the solution, I would be very grateful.
The detail of the problem was thus:
At the tee, X players flip discs to decide the order of play.
On each flip, if a single disc lands different to all the others, that player is 'out' i.e. next in order of play.
If no single disc lands differently (i.e. they all land the same, or a split of, for example, three 'up' and two 'down') then discs are reflipped.
When down to the last 2 players, one will call 'odd' (discs land differently) or 'even' (discs land the same) to decide the last positions.
Question:
What is the probability, given X players at the tee, that only X number of flips are required i.e. a player is 'out' on each flip.
So for 4 players only 4 flips are required (including the final odd/even flip).
Answers, on a virtual postacrd, to the forum...
Archive material (if you have any) to me, please.
Many thanks in advance.
BOF
Big Maths
many years ago (quite a big many), I set a probability teaser in one of the early BDGA newsletters.
I cannot find any copy of my original working, either on computers or on paper!
If anyone has an archive of my original problem and the solution, I would be very grateful.
The detail of the problem was thus:
At the tee, X players flip discs to decide the order of play.
On each flip, if a single disc lands different to all the others, that player is 'out' i.e. next in order of play.
If no single disc lands differently (i.e. they all land the same, or a split of, for example, three 'up' and two 'down') then discs are reflipped.
When down to the last 2 players, one will call 'odd' (discs land differently) or 'even' (discs land the same) to decide the last positions.
Question:
What is the probability, given X players at the tee, that only X number of flips are required i.e. a player is 'out' on each flip.
So for 4 players only 4 flips are required (including the final odd/even flip).
Answers, on a virtual postacrd, to the forum...
Archive material (if you have any) to me, please.
Many thanks in advance.
BOF
Big Maths
BDGA #33
PDGA #8835
http://www.ashvillediscgolf.co.uk
PDGA #8835
http://www.ashvillediscgolf.co.uk
Re: BOF's probability problem...
Surely if there are X players, only X1 flips would be required as the final flip would decide the fate of 2 players
Hyzer Cup Champions 09/10, 10/11, 11/12, 12/13
BDGA #357
PDGA #45315
BDGA #357
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 Posts: 578
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Re: BOF's probability problem...
Always answer the question set in the paper Mark.
Paul Holden
BDGA No. 307
PDGA No. 34662
BDGA No. 307
PDGA No. 34662

 Posts: 578
 Joined: Wed Mar 05, 2008 10:34 pm
 Location: York
Re: BOF's probability problem...
Joking aside I think it goes something like this...
The odds that for any one flip between Y players all but one disc land the same way up (assuming an equal chance for each way up) is:
(1/2)^(Y1) , where I am using ^ to represent "to the power of"
Beginning with X players, and only looking at the flips required whilst there are 3 or more players left, because the last flip will always decide the last two players, the cumulative odds are:
(1/2)^(X1+... ...+Xn) ,X > 2 ,n = X2
Evaluating the sum of the arithmetic series in the exponent gives us:
(n/2)(2X+(n1)(1))
(n/2)(2X+1n)
Substituting for n gives us:
((X2)/2)(2X+1X+2)
((X2)/2)(X+3)
So the final odds that the minimum number of throws, which Mark correctly identifies as X1, would be used is:
(1/2)^((X2)/2)(X+3)
Am I even close???
The odds that for any one flip between Y players all but one disc land the same way up (assuming an equal chance for each way up) is:
(1/2)^(Y1) , where I am using ^ to represent "to the power of"
Beginning with X players, and only looking at the flips required whilst there are 3 or more players left, because the last flip will always decide the last two players, the cumulative odds are:
(1/2)^(X1+... ...+Xn) ,X > 2 ,n = X2
Evaluating the sum of the arithmetic series in the exponent gives us:
(n/2)(2X+(n1)(1))
(n/2)(2X+1n)
Substituting for n gives us:
((X2)/2)(2X+1X+2)
((X2)/2)(X+3)
So the final odds that the minimum number of throws, which Mark correctly identifies as X1, would be used is:
(1/2)^((X2)/2)(X+3)
Am I even close???
Paul Holden
BDGA No. 307
PDGA No. 34662
BDGA No. 307
PDGA No. 34662

 Posts: 578
 Joined: Wed Mar 05, 2008 10:34 pm
 Location: York
Re: BOF's probability problem...
On the archive front the only BDGA newsletter I have is Issue 1 and it isn't in there I have nothing between that and Inflight 2008, nice pic of Chris OB on the front.
Paul Holden
BDGA No. 307
PDGA No. 34662
BDGA No. 307
PDGA No. 34662
Re: BOF's probability problem...
nice try, Paul, but I think it is simpler than that  unless your solution simplifies down to what I have in front of me here...
I have a nice solution just worked out, but am willing to give others a look before I reveal my solution (in case I'm right or wrong!).
I'll PM you my solution.
BOF
Big Pencil
I have a nice solution just worked out, but am willing to give others a look before I reveal my solution (in case I'm right or wrong!).
I'll PM you my solution.
BOF
Big Pencil
BDGA #33
PDGA #8835
http://www.ashvillediscgolf.co.uk
PDGA #8835
http://www.ashvillediscgolf.co.uk
Re: BOF's probability problem...
If only we could use the superscript BBCode 
Admin, if you feel like adding superscript and subscript functionality...
http://www.phpbbdevelopers.net/archive/ ... ?f=27&t=93
Thanks
B[sup]O[/sup]F
Admin, if you feel like adding superscript and subscript functionality...
http://www.phpbbdevelopers.net/archive/ ... ?f=27&t=93
Thanks
B[sup]O[/sup]F
BDGA #33
PDGA #8835
http://www.ashvillediscgolf.co.uk
PDGA #8835
http://www.ashvillediscgolf.co.uk
Re: BOF's probability problem...
Some of this maths is beyond me, but Paul, is your initial probability formula correct?
i.e. the probability that one of the discs is different being (1/2)^(Y1)
I don't know what the answer is, but if, say, three discs were flipped, there would be 6 ways out of 8 for one of the discs to be different, which is a probability of 3/4, but your formula would calculate it as 1/4.
The number of ways of one disc being different should be the 2nd number from each end of Pascal's triangle added together.
Then the probability is that number divided by the total in that row of the triangle, so for:
2 discs = 2/4 = 1/2
3 discs = 6/8 = 3/4
4 discs = 8/16 = 1/2
5 discs = 10/32 = 5/16
6 discs = 12/64 = 3/16
etc...
I'm not sure what formula gives that series though...
i.e. the probability that one of the discs is different being (1/2)^(Y1)
I don't know what the answer is, but if, say, three discs were flipped, there would be 6 ways out of 8 for one of the discs to be different, which is a probability of 3/4, but your formula would calculate it as 1/4.
The number of ways of one disc being different should be the 2nd number from each end of Pascal's triangle added together.
Then the probability is that number divided by the total in that row of the triangle, so for:
2 discs = 2/4 = 1/2
3 discs = 6/8 = 3/4
4 discs = 8/16 = 1/2
5 discs = 10/32 = 5/16
6 discs = 12/64 = 3/16
etc...
I'm not sure what formula gives that series though...
Tom
ND
ND
Re: BOF's probability problem...
At a glance: 2X/(2^X)LostMeow wrote: 2 discs = 2/4 = 1/2
3 discs = 6/8 = 3/4
4 discs = 8/16 = 1/2
5 discs = 10/32 = 5/16
6 discs = 12/64 = 3/16
etc...
I'm not sure what formula gives that series though...
[Standard post disclaimer] My posts are never intended to undermine the work of the Board or individuals putting in effort to grow the sport, they are my honest thoughts on the best ways to grow the game
BDGA: 145
PDGA: 8824
BDGA: 145
PDGA: 8824
Re: BOF's probability problem...
D'oh. Knew it would be deceptively simple! Anyway, is that the right probability, and can it be arithmetically progressed like Paul did with the other one?bruce wrote:
At a glance: 2X/(2^X)
Tom
ND
ND

 Posts: 578
 Joined: Wed Mar 05, 2008 10:34 pm
 Location: York
Re: BOF's probability problem...
Yeah, ignore that, I am home ill today and I think it was showing...LostMeow wrote:Some of this maths is beyond me, but Paul, is your initial probability formula correct?
i.e. the probability that one of the discs is different being (1/2)^(Y1)
I don't know what the answer is, but if, say, three discs were flipped, there would be 6 ways out of 8 for one of the discs to be different, which is a probability of 3/4, but your formula would calculate it as 1/4.
Paul Holden
BDGA No. 307
PDGA No. 34662
BDGA No. 307
PDGA No. 34662

 Posts: 578
 Joined: Wed Mar 05, 2008 10:34 pm
 Location: York
Re: BOF's probability problem...
I am out by factor of Y...
(1/2)^(Y1)
((1/2)^(1))((1/2)^(Y))
2((1/2)^(Y)
2/(2^Y)
Whereas Bruce has:
2Y/(2^Y)
(1/2)^(Y1)
((1/2)^(1))((1/2)^(Y))
2((1/2)^(Y)
2/(2^Y)
Whereas Bruce has:
2Y/(2^Y)
Paul Holden
BDGA No. 307
PDGA No. 34662
BDGA No. 307
PDGA No. 34662
Re: BOF's probability problem...
Is this the answer?
http://www.wolframalpha.com/input/?i=pr ... rom+3+to+n
Formula is:
(2^(0.5(n1)n))n!
http://www.wolframalpha.com/input/?i=pr ... rom+3+to+n
Formula is:
(2^(0.5(n1)n))n!
Tom
ND
ND

 Posts: 578
 Joined: Wed Mar 05, 2008 10:34 pm
 Location: York
Re: BOF's probability problem...
Looks good, I had just got to:
N! ( (1/2)^( (1/2)((N^2)N) )
Which I think is the same?
N! ( (1/2)^( (1/2)((N^2)N) )
Which I think is the same?
Paul Holden
BDGA No. 307
PDGA No. 34662
BDGA No. 307
PDGA No. 34662
Re: BOF's probability problem...
LostMeow, Paul  well done!
I wrote it as:
n!/(2^(0.5n(n1))
Ten points to Gryffindor!
Class dismissed.
BOF
Big Calculator
I wrote it as:
n!/(2^(0.5n(n1))
Ten points to Gryffindor!
Class dismissed.
BOF
Big Calculator
BDGA #33
PDGA #8835
http://www.ashvillediscgolf.co.uk
PDGA #8835
http://www.ashvillediscgolf.co.uk